The implication (i)\(\Rightarrow \) (ii) is covered by item (1) of Theorem REF , the implication (ii)\(\Rightarrow \) (iii) is evident, and the implication (i)\(\Rightarrow \) (ii) follows from the item (4) of Theorem REF . It remains to demonstrate that (iv)\(\Rightarrow \) (i). This fact is well-known ([1]}) and may be deduced from an analogous theorem for uniform spaces. Nevertheless, for the reader's convenience (and for a reference below) we prefer to give a direct proof. So, let \(U_n \in \mathfrak {N}_0\) , \(U_1 \supset U_2 \supset \ldots \)  be a base of neighborhoods of zero with the property that \(U_{n+1} + U_{n+1} \subset U_n\) , and let \(\mathfrak {F}\)  be a Cauchy filter on \(X\) . For each \(n \in {\mathbb {N}}\)  pick \(A_n \in \mathfrak {F}\)  such that \(A_n - A_n \subset U_n\) , and select an \(x_n \in \bigcap _{k=1}^n A_k\) . Then \(x = (x_n)\)  is a Cauchy sequence. Indeed, for every \(U \in \mathfrak {N}_0\)  there is an \(N \in {\mathbb {N}}\)  such that \(U_N \subset U\) . Then, for \(n, m \geqslant N\)  we have that \(x_n - x_m \in A_N - A_N \subset U_N \subset U\) .
Since \(x = (x_n)\)  is a Cauchy sequence and \(X\)  is sequentially complete, there is \(y:= \lim _{n \rightarrow \infty } x_n\) . Let us show that the same \(y\)  is the limit of \(\mathfrak {F}\) . Consider an arbitrary neighborhood \(V \in \mathfrak {N}_y\) . \(V - y \in \mathfrak {N}_0\) , so there is \(m \in {\mathbb {N}}\)  such that \(U_{m} \subset V - y\) . By the definition of \(y\) , there is \(k > m\)  such that \(x_k \in U_{m+1} + y\) . For this \(k\)  we have \(A_k - x_k \subset A_k - A_k \subset U_k \subset U_{m+1}\) , consequently \(A_k \subset U_{m+1} + x_k\) . This means that
\(V \supset U_m + y \supset U_{m+1} + U_{m+1} + y \supset U_{m+1} + x_k \supset A_k,\) 
so \(V \in \mathfrak {F}\) .
